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Math Expert V
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A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

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Math Expert V
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Posts: 7988
A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

Equation - 64 classes and 37 teachers...

LEAST possible
- all have two so 37*2 = 74, more than 64 classes, so it is possible that NONE of classes had 3 teachers..

GREATEST possible - say x classes have 3 teacher..... so 3x + (37-x) = 64............3x+37-x = 64...............2x = 27................x = 13.5.....
take the INTEGER value as greatest possible = 13

ans 0 and 13..
A
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

program was tried out in 2 classes in 32 elementary school. So total number of classes are 2*32= 64 classes.

Teachers involved= 37.

Each teacher took at least 1 class; that means that 37 classes are taken and remaining classes 64-37= 27 classes are divided among 37 teachers again.

If 1 class is given to each teacher, 27 can be divided among 27 teachers, making no teacher teaching 3 classes.

Hence, least possible number is 0.

But if 2 classes are given to any of teh 37 teachers, we can distribute the classes among 27/2= 13.5= 13 teachers.

A is the answer
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Marshall & McDonough Moderator D
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A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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x --> Number of teachers who teach in 3 subjects
y --> Number of teachers who teach in 2 subjects
z --> Number of teachers who teach in 1 subject

Total number of classes = 2 * 32 = 64

x + y + z = 37
3x + 2y + z = 64

Check the possibilities from the answer choices:
Put x = 0 --> y + z = 37 and 2y + z = 64 --> y = 27 and z = 10 (Possible minimum)

Put x = 14 --> y + z = 23 and 2y + z = 22 --> y = -1 (Not possible)

Put x = 13 --> y + z = 24 and 2y + z = 25 --> y = 1 and z = 23 (Possible maximum)

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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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chetan2u wrote:
Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

Equation - 64 classes and 37 teachers...

LEAST possible
- all have two so 37*2 = 74, more than 64 classes, so it is possible that NONE of classes had 3 teachers..

GREATEST possible - say x classes have 3 teacher..... so 3x - (37-x) = 64............3x-37+x = 64...............2x = 27................x = 13.5.....
take the INTEGER value as greatest possible = 13

ans 0 and 13..
A

Hi Chetan,

Pretty cool technique! Thanks.

For the greatest possible value, the equation should be 3x + (37-x) = 64, right?
Math Expert V
Joined: 02 Aug 2009
Posts: 7988
Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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jayshah0621 wrote:
chetan2u wrote:
Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

Equation - 64 classes and 37 teachers...

LEAST possible
- all have two so 37*2 = 74, more than 64 classes, so it is possible that NONE of classes had 3 teachers..

GREATEST possible - say x classes have 3 teacher..... so 3x - (37-x) = 64............3x-37+x = 64...............2x = 27................x = 13.5.....
take the INTEGER value as greatest possible = 13

ans 0 and 13..
A

Hi Chetan,

Pretty cool technique! Thanks.

For the greatest possible value, the equation should be 3x + (37-x) = 64, right?

Thanks .. typo error.. edited
kudos
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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11
There is also a shortcut in the answers.
One may notice that greatest possible values differ in each answer choice in contrast to the least values, which repeat.
To find out the greatest value you should count the total classes (32*2=64), then subtract the total #of teachers since we know from the question that each teacher taught at least one class (64-37=27). Thus we get a number of the available extra-classes for teachers, and all that we need is just to count how many teachers could take 2 more classes, which is 27/2 = 13.5. So the greatest possible value of the # of teachers who had 3 classes is 13.
Only answer A has this option.
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A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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3
a+2b+3c=64
a+b+c =37

subtracting second equation from first we obtain b+2c = 27
or b = 27 - 2c. This shows c cannot be more than 13, otherwise b would be negative. C can be 0 as well.
Below is a table of values depicting the range.

a---- -b----c
10--- 27---0
23--- -1---13

Originally posted by Henry S. Hudson Jr. on 31 Jul 2016, 10:01.
Last edited by Henry S. Hudson Jr. on 25 Mar 2019, 16:37, edited 2 times in total.
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Joined: 22 Jul 2016
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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MannyOliveira wrote:
There is also a shortcut in the answers.
One may notice that greatest possible values differ in each answer choice in contrast to the least values, which repeat.
To find out the greatest value you should count the total classes (32*2=64), then subtract the total #of teachers since we know from the question that each teacher taught at least one class (64-37=27). Thus we get a number of the available extra-classes for teachers, and all that we need is just to count how many teachers could take 2 more classes, which is 27/2 = 13.5. So the greatest possible value of the # of teachers who had 3 classes is 13.
Only answer A has this option.

Money shot right here.
Manager  S
Joined: 03 Jan 2017
Posts: 134
Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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tuff question:

let's first analyse for min, there are 32*2=64 classes total
there are 37 teachers*2=74. this means that no teacher may had 3 lessons
so 0 is min

lets analyze for max: we have x teachers that toguht 3 lessons 3*x+other teachers*classes=64
other teachers are 37-x
3*x+37-x=64
2x=27
x=13,3
as we look for an integer 13 is the max value
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

We are given that a certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Thus, there were a total of 2 x 32 = 64 classes under this program.

If we let a = the number of teachers teaching one class, b = the number of teachers teaching two classes, and c = the number of teachers teaching 3 classes, we can create the following equations:

a + b + c = 37

a + 2b + 3c = 64

Subtracting equation 1 from equation 2, we have:

(a + 2b + 3c = 64) - (a + b + c = 37)

b + 2c = 27

2c = 27 - b

c = (27 - b)/2

We see that c is the GREATEST when b = 1, and thus (27 - 1)/2 = 26/2 = 13.

We also see that c is the LEAST when b = 27, and thus (27 - 27)/2 = 0/2 = 0.

So, the range of values of n is 0 to 13.

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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

This is my 2 cents.
I did back-solving and started with B because the q asks for least and greatest.

if n = 14, then 14 teachers cover 42 classes.
Remaining classes are 64-42 = 22
Remaining teachers are 37-14 = 23
As the stem says each teach at least covers 1 teacher, this is not good.

So i moved on to A.

if n=13, then 13 teachers cover 39 classes.
Remaining classes are 64-39 = 25
Remaining teachers are 37-13 = 24, which is good as we have more classes than teachers.
if n=0, then we have 37 teachers cover 64 classes.
As these teachers can teach anywhere between 1-2, we know that we have enough teachers to cover all classes without any teacher teaching 3 classes.

Hence A
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GMAT 1: 780 Q51 V46 Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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Here are a couple of video explanations for this question:

Algebra:

Backsolving:

Intern  B
Joined: 14 May 2015
Posts: 47
Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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total class=32*2=64
total teacher=37
least:
distribute 1 class each to 37 teachers. remain=64-37=27
again distribute 1 each class to 27 teachers.
only 27 teachers gets 2 classes. no one gets 3 classes.
least value of p=0
greatest:
distribute 1 class each to 37 teachers. remain=64-37=27
again distribute 2 classes each to 27/2=13 teachers. and 1 class to other teachers.
13 teachers get 3 classes.
max value of p=13
ans: A) 0 and 13
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

I see this question as a variant of the must be/could be true variant with some element of data sufficiency and ultimately a ratio problem. One of A and B is sufficient to answer the question as 3 class teachers could be zero since there the average number of class per teacher, i.e 64/37 is less than 2. Next is to plug in 13 and 14, if we want the maximum number of 3 the maximum number of 3 class teachers will need the maximum number of 1 class teachers. Now we either have to distribute 25 or 22 remaining classes among 24 or 23 teachers respectively. Obviously the answer is 25 classes to the remaining 24 teachers since every teacher teaches a class at least, corresponding to the values 39:2:23 for the ratio 3:2:1. Hence A.
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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Henry S. Hudson Jr. wrote:
a+2b+3c=64
a+b+c =37

subtracting second equation from first we obtain b+2c = 27
or b = 27 - 2c. This shows c cannot be more than 13, otherwise b would be negative. C can be 0 as well.
Below is a table of values depicting the range.

a b c
10 27 0
23 1 13

Thanks, what signifies you to SUBTRACT the 2nd equation from the 1st versus ADDing them together?
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

Check the answer choices (ALWAYS check the answer choices before choosing a particular solution strategy)
I see that, for each answer choice, the second value (the greatest value of n) is different. So, let's test some of these values.

Let's start by testing answer choice B (0 and 14)
I'd like to start here, since we're asked to identify the greatest value of n, and answer choice B has the biggest possible value of n.
So, is it possible to have 14 teachers who teach 3 classes?
Well, (14)(3) = 42 classes
There are 64 classes altogether (2 classes in each of the 32 schools, means a total of 64 classes)
So, the number of classes that still require teachers = 64 - 42 = 22

How many teachers are remaining?
So far, 14 of the 37 teachers are accounted for (they're the ones who are teaching 3 classes each)
So, the number of teachers remaining = 37 - 14 = 23
Can these 23 remaining teachers cover the remaining 22 classes?
NO!
Each teacher must teach AT LEAST ONE class. So, there aren't enough classes needed for each teacher to teach at least one class.
So, we can ELIMINATE answer choice B.

IMPORTANT: We were VERY CLOSE with answer choice B. We were just one class short of meeting our goal. So, I am quite confident that the greatest possible values of n is 13 (answer choice A). Let's find out.

We'll test answer choice A (0 and 13)
Well, (13)(3) = 39 classes
There are 64 classes altogether
So, the number of classes that still require teachers = 64 - 39 = 25

So far, 13 of the 37 teachers are accounted for. So, the number of teachers remaining = 37 - 13 = 24
Can these 24 remaining teachers cover the remaining 25 classes?
YES!
23 of the teachers can teach 1 class each, and the other teacher can teach 2 classes.
Since the greatest possible value of n is 13, the correct answer is ...

Cheers,
Brent
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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I will be utilizing a low level approach to solve this question. Such an approach request we don't get into any messy algebra and rely on mere arithmetic and logic:

total classes: 32 * 2 = 64
total teachers: 37

After all 37 teachers have taken 1 class, we are left with 27 classes to be taught.

Those 27 classes could be taught by 27 teachers, which results in following:
teachers taking 1 class: 37
teachers taking 2 classes: 27

teachers taking 3 classes: 0 (minimum value of n) --> this means we are now left with option A and B

the remaining 27 classes if taught by minimum number of teachers will maximize the number teachers, who took 3 classes, so let us give 2 more classes to each of the teachers. This means that 26 classes could be taught by 13 teachers, each of whose total classes will become 3. We are now left with 1 class. So the new numbers are:
teachers taking 1 class: 37
teachers taking 2 classes: 1

teachers taking 3 classes: 13 (maximum value of n) --> this means that our answer is A
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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I solved it this way:

We have 32 Schools, that means we have 64 classes.
Furthermore we have 37 teachers.
Now assume that the teachers are lazy and no one will do more than neccessary. So every teacher takes one class. Than there are 27(64-37) classes left without a teacher. So 27 poor teacher have to teach 2 classes. The rest of them can stay lazy and teach one. --> 0 teacher teach 3 classes.

But if the teachers are very engaged, the remaining 27 classes are devided between the teachers in a way that receives the highest possible number of teachers that teach 3 classes(consider that every teacher have to teach at least one class). So you devide 27:2 = 13 +1
--> 13 teachers will get 2 classes more so they teach 3 classes. The remaining class is teached by a teacher who teach 2 classes.

Emagine you have a bucket with 27 apples. and a few hundred kids, everybody already have an apple but everybody also wants 2 more. What can you do? you can give 13 children 2 apples more.

Hope it helps
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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THIS is hard

we have 32x2=64 classses

suppose each teacher teach one class, we have 37 classeses. and
64-37=27
so there is 27 classeses left. if 27 teacher each take one class more, then there is no teacher who take 3 classess. answer can be 1 or 2
if some teachers get two classes more, we have 13 teacher who take 2 classes more (total they get 3 classese) and one classe is left. so one teacher take one class more.
we can not have 14 teacher get 2 classese more because there is only 27, not 28 classese left. Re: A certain experimental mathematics program was tried out in 2 classes   [#permalink] 03 Dec 2017, 09:38

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